1967. Number of Strings That Appear as Substrings in Word

Description

Given an array of strings patterns and a string word, return the number of strings in patterns that exist as a substring in word.

A substring is a contiguous sequence of characters within a string.

Example 1:

Input: patterns = ["a","abc","bc","d"], word = "abc"
Output: 3
Explanation:
- "a" appears as a substring in "abc".
- "abc" appears as a substring in "abc".
- "bc" appears as a substring in "abc".
- "d" does not appear as a substring in "abc".
3 of the strings in patterns appear as a substring in word.

Example 2:

Input: patterns = ["a","b","c"], word = "aaaaabbbbb"
Output: 2
Explanation:
- "a" appears as a substring in "aaaaabbbbb".
- "b" appears as a substring in "aaaaabbbbb".
- "c" does not appear as a substring in "aaaaabbbbb".
2 of the strings in patterns appear as a substring in word.

Example 3:

Input: patterns = ["a","a","a"], word = "ab"
Output: 3
Explanation: Each of the patterns appears as a substring in word "ab".

Constraints:

• 1 <= patterns.length <= 100
• 1 <= patterns[i].length <= 100
• 1 <= word.length <= 100
• patterns[i] and word consist of lowercase English letters.

Solutions

Solution 1

1
2
3

class Solution:
    def numOfStrings(self, patterns: List[str], word: str) -> int:
        return sum(p in word for p in patterns)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

class Solution {
    public int numOfStrings(String[] patterns, String word) {
        int ans = 0;
        for (String p : patterns) {
            if (word.contains(p)) {
                ++ans;
            }
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10

class Solution {
public:
    int numOfStrings(vector<string>& patterns, string word) {
        int ans = 0;
        for (auto& p : patterns) {
            ans += word.find(p) != string::npos;
        }
        return ans;
    }
};

1
2
3
4
5
6
7
8

func numOfStrings(patterns []string, word string) (ans int) {
	for _, p := range patterns {
		if strings.Contains(word, p) {
			ans++
		}
	}
	return
}

1
2
3
4
5
6
7
8
9

function numOfStrings(patterns: string[], word: string): number {
    let ans = 0;
    for (const p of patterns) {
        if (word.includes(p)) {
            ++ans;
        }
    }
    return ans;
}